Q96P - University Physics with Modern Physics

Question

A camera with a 90-mm-focal-length lens is focused on an object 1.30 m from the lens. To refocus on an object 6.50 m from the lens, by how much must the distance between the lens and the sensor be changed? To refocus on the more distant object, is the lens moved toward or away from the sensor?

Step-by-Step Solution

Answer

The distance is changed by 5.4mm towards the film to refocus an object 6.5m from the lens.

Step 1:Relationship between object and image distance and height.

Object’s distance is s and image distance is s’.

The relationship is $\frac{1}{s}+\frac{1}{s\text{'}}=\frac{1}{\mathit{f}}$ 

Object’s height is y and image height is y’,

Therefore $y\text{'}=-\frac{s\text{'}}{s}y$ 

Step 2:Calculation of the required change in the distance of the lens.

Using the relationship of distance between object and the image.

The first object

$$\begin{gathered} s{\text{'}}_1=\frac{\mathbf{s}\mathbf{f}}{\mathbf{s}\mathbf{-}\mathbf{f}} \\ =\frac{1300*\left(90\right)}{1300-90)} \\ =96.7\mathit{c}\mathit{m} \end{gathered}$$

 

The second object

$$\begin{gathered} s{\text{'}}_2=\frac{\mathbf{sf}}{\mathbf{s}\mathbf{-}\mathbf{f}} \\ =\frac{6500*\left(90\right)}{6500-90} \\ =91.3\mathit{c}\mathit{m} \end{gathered}$$

 

Therefore the distance that is to be changed will be,

$$\begin{gathered} s{\text{'}}_1-s{\text{'}}_2=96.7-91.3 \\ =5.4\mathit{m}\mathit{m} \end{gathered}$$